Map Life-Table Entries to Valuation Probabilities

From Life-Table Entries to Contract Probabilities

A life table records mortality by age, while a policy specifies benefits by policy year or duration. To value the policy, first translate the table into probabilities for the exact interval and conditioning age relevant to each cash flow. The key question is often not just whether someone dies, but from which age, over which interval, and conditional on being alive when?

This tutorial uses integer-age life-table entries. It assumes the life table and the contract use the same mortality basis, and that the policy is issued at an integer age. Fractional-age interpolation is not needed for these calculations. Once the relevant probabilities are mapped, they can be used with the interest and benefit-timing conventions of the contract.

Definition: In a life table, $l_x$ is the number of lives surviving to exact age $x$, and $d_x$ is the number dying between exact ages $x$ and $x+1$. Thus $d_x=l_x-l_{x+1}$. The one-year death probability is $q_x=d_x/l_x$, and the one-year survival probability is $p_x=l_{x+1}/l_x=1-q_x$, conditional on survival to age $x$.

The life-table count $l_x$ need not be an actual population count. It may be a theoretical cohort count, often called a radix. Probabilities depend on ratios of entries, so multiplying every $l_x$ by the same positive constant does not change a probability.

Mapping One-Year Entries to Longer Intervals

A life-table $q_x$ is conditional on being alive at age $x$. By contrast, a contract issued at age $x$ may need the probability of surviving several years, or dying in a particular later policy year. These probabilities use successive life-table entries.

Theorem: For integer $n\geq 0$, the probability that a life aged $x$ survives $n$ years is ${}_np_x=l_{x+n}/l_x$. The probability of dying within those $n$ years is ${}_nq_x=1-{}_np_x$. The probability of dying during policy year $k$, where year 1 is $(x,x+1]$, is ${}_{k-1|}q_x={}_{k-1}p_xq_{x+k-1}=(l_{x+k-1}-l_{x+k})/l_x$.

The denominator $l_x$ in these expressions represents the cohort alive at issue age $x$. If instead the probability is conditional on reaching age $x+m$, use $l_{x+m}$ as the denominator. In particular, the probability of surviving from age $x+m$ to age $x+n$, for $n\geq m$, is $l_{x+n}/l_{x+m}$; the probability of dying in that interval conditional on reaching age $x+m$ is $1-l_{x+n}/l_{x+m}$.

The multi-year survival formula follows by multiplying conditional one-year survival probabilities. The factors telescope:

{}_np_x=p_xp_{x+1}\cdots p_{x+n-1} =\frac{l_{x+1}}{l_x}\frac{l_{x+2}}{l_{x+1}}\cdots\frac{l_{x+n}}{l_{x+n-1}} =\frac{l_{x+n}}{l_x}.

For death in policy year $k$, the life must first survive $k-1$ years and then die in the next year. This gives ${}_{k-1}p_xq_{x+k-1}$. The life-table count difference $l_{x+k-1}-l_{x+k}$ counts deaths in that year, so dividing it by the issue cohort $l_x$ gives the same result.

Fix the conditioning age.
Use the age at issue if the probability is measured from issue; use the attained age if the contract interval begins later.

Identify the interval.
For survival, locate the life-table entries at the start and end ages. For death in one year, take the difference between adjacent entries.

Choose the denominator.
Divide by the number alive at the age on which the probability is conditional.

Match the probability to the cash flow.
Discount a benefit to its payment time, and use a survival probability for each premium or annuity payment that is contingent on being alive.

Worked Valuations

The examples state the mortality and interest assumptions explicitly. For discrete insurance benefits, the benefit is paid at the end of the year of death. For an annuity-due premium, a payment is made at the start of each policy year if the insured is then alive.

Worked Example: Three-Year Term Insurance from Life-Table Counts

A hypothetical insurer issues a three-year term policy at age 60. It pays \$20,000 at the end of the year of death, if death occurs during the term. The effective annual interest rate is $i=5\%$. The life table has $l_{60}=10{,}000$, $l_{61}=9{,}600$, $l_{62}=9{,}120$, and $l_{63}=8{,}664$. Find the expected present value of the benefit.

Method. First find each policy-year death probability using the issue cohort as denominator. Then discount each possible end-of-year benefit. This avoids treating the one-year $q_x$ values as if they were already measured from age 60.

The one-year probabilities are $q_{60}=(10{,}000-9{,}600)/10{,}000=0.04$, $q_{61}=(9{,}600-9{,}120)/9{,}600=0.05$, and $q_{62}=(9{,}120-8{,}664)/9{,}120=0.05$. The probabilities of death in the three policy years, each measured from issue, are therefore

\begin{aligned} P(\text{death in year 1})&=\frac{10{,}000-9{,}600}{10{,}000}=0.0400,\\ P(\text{death in year 2})&=\frac{9{,}600-9{,}120}{10{,}000}=0.0480,\\ P(\text{death in year 3})&=\frac{9{,}120-8{,}664}{10{,}000}=0.0456. \end{aligned}

These probabilities sum to $0.1336$, which equals $1-l_{63}/l_{60}=1-0.8664$, a useful check. With $v=1/1.05$, the expected present value is

\begin{aligned} \operatorname{EPV} &=20{,}000\left(0.0400v+0.0480v^2+0.0456v^3\right)\\ &=20{,}000\left(\frac{0.0400}{1.05} +\frac{0.0480}{1.1025} +\frac{0.0456}{1.157625}\right)\\ &=20{,}000(0.0380952381+0.0435374150+0.0393909910)\\ &=2{,}420.4729\ldots \end{aligned}

The expected present value is \$2,420.47, rounded to the nearest cent. The final check is that the undiscounted expected benefit would be \$20,000 times $0.1336$, or \$2,672; discounting benefits paid in the future should reduce that amount.

Worked Example: A Deferred Term Benefit

A hypothetical policy is issued at age 50. It pays \$10,000 at the end of the year of death if death occurs in policy year 3 or policy year 4, and pays nothing otherwise. The effective annual interest rate is $i=4\%$. The table gives $l_{50}=1{,}000{,}000$, $l_{51}=980{,}000$, $l_{52}=955{,}500$, $l_{53}=917{,}280$, and $l_{54}=871{,}416$. Find the expected present value.

Method. The coverage is deferred for two years, so the relevant deaths are the differences $l_{52}-l_{53}$ and $l_{53}-l_{54}$. Divide each by $l_{50}$, since the requested probabilities are for a life at issue, not just for a life that has already reached age 52.

The first covered-year death probability is $(955{,}500-917{,}280)/1{,}000{,}000=0.038220$. The second is $(917{,}280-871{,}416)/1{,}000{,}000=0.045864$. Equivalently, survival to age 52 is $955{,}500/1{,}000{,}000=0.9555$; the conditional death probabilities at ages 52 and 53 are $0.04$ and $0.05$, giving $0.9555(0.04)=0.03822$ and $0.9555(0.96)(0.05)=0.045864$.

Benefits are paid at the ends of years 3 and 4. Thus

\begin{aligned} \operatorname{EPV} &=10{,}000\left(\frac{0.038220}{1.04^3} +\frac{0.045864}{1.04^4}\right)\\ &=10{,}000(0.0339774409+0.0392047393)\\ &=731.8218\ldots \end{aligned}

The expected present value is \$731.82, rounded to the nearest cent. As a probability check, the covered death probabilities sum to $0.084084$, which also equals $(l_{52}-l_{54})/l_{50}$.

Worked Example: Use Life-Table Probabilities in a Premium and Reserve

A hypothetical two-year term policy issued at age 40 pays \$10,000 at the end of the year of death. The effective annual interest rate is $i=5\%$. The life table has $l_{40}=20{,}000$, $l_{41}=19{,}600$, and $l_{42}=19{,}012$. Level net premiums are payable at the beginning of each policy year while the insured is alive. Find the annual net premium, then find the prospective reserve at time 1 immediately after the second premium has been paid, conditional on survival to age 41.

Method. Use the issue-based death probabilities for the benefit EPV and the survival probability to time 1 for the second premium. At time 1, immediately after that premium, no future premiums remain; the reserve is the EPV of the remaining benefit conditional on survival to age 41.

The one-year probabilities are $q_{40}=400/20{,}000=0.02$, $p_{40}=0.98$, and $q_{41}=588/19{,}600=0.03$. Death in policy year 2, measured from issue, has probability $p_{40}q_{41}=0.98(0.03)=0.0294$. The benefit EPV is

10{,}000\left(\frac{0.02}{1.05}+\frac{0.0294}{1.05^2}\right) =10{,}000(0.0190476190+0.0266666667) =457.142857\ldots.

The EPV of the two premium payments per policy issued is $P(1+0.98/1.05)$: the first is certain at issue, and the second is paid at time 1 only if the insured survives the first year. Since $1+0.98/1.05=1.933333\ldots$, equivalence gives

P=\frac{457.142857\ldots}{1.933333\ldots} =236.4532\ldots.

The annual net premium is \$236.45, rounded to the nearest cent. At time 1, conditional on survival, the probability of death in the remaining year is $q_{41}=0.03$. The remaining benefit is payable at time 2, one year after the reserve valuation date, and the second premium has already been paid. Therefore

{}_1V_{40}=10{,}000\frac{0.03}{1.05}=\$285.714285\ldots=\$285.71.

The reserve is \$285.71, rounded to the nearest cent. This is a prospective reserve immediately after the premium; valuing immediately before that premium would require including it as a future premium.

Common Traps and Checks

A correct life-table calculation can still produce a wrong valuation if the probability and the cash flow refer to different conditions or times. Use these checks before finalizing a calculation:

Using the wrong denominator. $q_{x+1}$ is conditional on reaching age $x+1$. Death in the second year from issue has probability $p_xq_{x+1}$, not $q_{x+1}$ alone.
Confusing conditional and issue-based probabilities. The probability of death during ages $x+m$ to $x+n$, conditional on reaching $x+m$, is $(l_{x+m}-l_{x+n})/l_{x+m}$. From issue age $x$, it is $(l_{x+m}-l_{x+n})/l_x$.
Discounting to the wrong time. If a death benefit is paid at the end of policy year $k$, discount it by $v^k$. A benefit payable immediately at death would have different timing.
Misreading a deferred interval. A two-year deferral followed by coverage beginning in year 3 does not include deaths in years 1 or 2. Use the life-table differences for the covered years.
Ignoring the payment condition on premiums. A premium due at the start of a year is included only if the insured is alive at that time. State whether the valuation is immediately before or after a premium.

As basic consistency checks, life-table counts should not increase with age in an ordinary mortality table; $p_x+q_x=1$; and probabilities for disjoint death years should sum to the probability of death over their combined interval. At an integer age, the ratios of $l_x$ entries provide the direct mapping; do not substitute a rate conditional on a later age for a probability measured from issue.

Check Your Understanding

For each problem, identify the conditioning age and interval before calculating. Round monetary answers to the nearest cent.

A table has $l_{30}=50{,}000$ and $l_{31}=49{,}000$. Find $q_{30}$ and $p_{30}$. (Answer: $q_{30}=0.0200,\ p_{30}=0.9800$.)
A table has $l_{50}=120{,}000$, $l_{51}=117{,}600$, and $l_{52}=114{,}072$. Find the probability of surviving two years from age 50 and the probability of dying within two years. (Answer: ${}_2p_{50}=0.9506,\ {}_2q_{50}=0.0494$.)
A contract issued at age 40 covers death in policy years 2 and 3. The table has $l_{40}=1{,}000{,}000$, $l_{41}=980{,}000$, $l_{42}=950{,}600$, and $l_{43}=912{,}576$. Find the issue-based probability of death in each covered year and survival through year 3. (Answer: year 2 $=0.0294$, year 3 $=0.038024$, and ${}_3p_{40}=0.912576$.)
A two-year term benefit of \$5,000 is paid at the end of the year of death. The effective annual interest rate is $i=5\%$, with $l_{60}=10{,}000$, $l_{61}=9{,}800$, and $l_{62}=9{,}506$. Find its expected present value. (Answer: $\$228.57$.)
A one-year term policy pays \$12,000 at the end of the year of death. At age 70, $l_{70}=8{,}000$ and $l_{71}=7{,}840$. The effective annual interest rate is $i=4\%$. A single net premium is paid at issue. Find the premium. (Answer: $12{,}000(0.02/1.04)=\$230.77$.)

What You'll Learn